Answer
$$s = 2\pi $$
Work Step by Step
$$\eqalign{
& y = \sqrt {16 - {x^2}} {\text{ over the interval }}\left[ {0,4} \right] \cr
& \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {\sqrt {16 - {x^2}} } \right] \cr
& \frac{{dy}}{{dx}} = \frac{{ - 2x}}{{2\sqrt {16 - {x^2}} }} \cr
& \frac{{dy}}{{dx}} = - \frac{x}{{\sqrt {16 - {x^2}} }} \cr
& {\text{The arc length of }}y{\text{ between }}\left[ {a,b} \right]{\text{ is}} \cr
& s = \int_a^b {\sqrt {1 + {{\left( {\frac{{dy}}{{dx}}} \right)}^2}} dx} {\text{, then}} \cr
& s = \int_0^4 {\sqrt {1 + {{\left( { - \frac{x}{{\sqrt {16 - {x^2}} }}} \right)}^2}} dx} \cr
& s = \int_0^4 {\sqrt {1 + \frac{{{x^2}}}{{16 - {x^2}}}} dx} \cr
& s = \int_0^4 {\sqrt {\frac{{16 - {x^2} + {x^2}}}{{16 - {x^2}}}} dx} \cr
& s = \int_0^4 {\frac{4}{{\sqrt {16 - {x^2}} }}dx} \cr
& s = \mathop {\lim }\limits_{a \to {4^ - }} \int_0^a {\frac{4}{{\sqrt {16 - {x^2}} }}dx} \cr
& {\text{Integrating}} \cr
& s = 4\mathop {\lim }\limits_{a \to {4^ - }} \left[ {{{\sin }^{ - 1}}\left( {\frac{x}{4}} \right)} \right]_0^a \cr
& s = 4\mathop {\lim }\limits_{a \to {4^ - }} \left[ {{{\sin }^{ - 1}}\left( {\frac{a}{4}} \right) - {{\sin }^{ - 1}}\left( {\frac{0}{4}} \right)} \right] \cr
& {\text{Evaluate the limit when }}a \to {4^ - } \cr
& s = 4\left( {\frac{\pi }{2} - 0} \right) \cr
& s = 2\pi \cr} $$
